Problem: Let $f(x)=\dfrac{3x^2+1}{x+2}$. Find $f'(-3)$. Choose 1 answer: Choose 1 answer: (Choice A) A $-10$ (Choice B) B $10$ (Choice C) C $28$ (Choice D) D $-28$
Explanation: Let's first find the derivative of $f$, i.e. the expression for $f'(x)$. Then, we can plug $x=-3$, into $f'(x)$ and evaluate. $f$ is a rational function. To find the derivative of rational functions, we use the quotient rule : $\begin{aligned} \dfrac{d}{dx}\left[\dfrac{u(x)}{v(x)}\right]&=\dfrac{\dfrac{d}{dx}[u(x)]v(x)-u(x)\dfrac{d}{dx}[v(x)]}{[v(x)]^2} \\\\ &=\dfrac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2} \end{aligned}$ Let's differentiate! = f ′ ( x ) = d d x ( 3 x 2 + 1 x + 2 ) = ( x + 2 ) d d x ( 3 x 2 + 1 ) − ( 3 x 2 + 1 ) d d x ( x + 2 ) ( x + 2 ) 2 = ( x + 2 ) ( 6 x ) − ( 3 x 2 + 1 ) ( 1 ) ( x + 2 ) 2 = 6 x 2 + 12 x − 3 x 2 − 1 ( x + 2 ) 2 = 3 x 2 + 12 x − 1 ( x + 2 ) 2 The quotient rule Differentiate ( 3 x 2 + 1 ) & ( x + 2 ) Expand \begin{aligned} &\phantom{=}f'(x) \\\\ &=\dfrac{d}{dx}\left(\dfrac{3x^2+1}{x+2}\right) \\\\ &=\dfrac{(x+2)\dfrac{d}{dx}(3x^2+1)-(3x^2+1)\dfrac{d}{dx}(x+2)}{(x+2)^2}&&\gray{\text{The quotient rule}} \\\\ &=\dfrac{(x+2)(6x)-(3x^2+1)(1)}{(x+2)^2}&&\gray{\text{Differentiate }(3x^2+1)\text{ & }(x+2)} \\\\ &=\dfrac{6x^2+12x-3x^2-1}{(x+2)^2}&&\gray{\text{Expand}} \\\\ &=\dfrac{3x^2+12x-1}{(x+2)^2} \end{aligned} So we found that $f'(x)=\dfrac{3x^2+12x-1}{(x+2)^2}$. Now let's plug $x= {-3}$ into $f'$ : $\begin{aligned} &\phantom{=}f'( {-3}) \\\\ &=\dfrac{3( {-3})^2+12({ -3})-1}{(( {-3})+2)^2} \\\\ &=\dfrac{27-36-1}{(-1)^2} \\\\ &=-10 \end{aligned}$ In conclusion, $f'(-3)=-10$.